Re: Multiple Voltage Sources?



John Popelish wrote:

The point to remember is that no circuit voltage exists in isolation,
except for floating charges. All voltages are the difference in
potential between two nodes. So, for a 5 volt signal from one source
to be connected to a device powered from a different source, both
sources have to be connected at one point, at least. That point would
be the negative sides of those two sources, in this case.

Aha, that seems to make sense with everything I've read, if I
understand it properly. Basically I'd have to connect the ground of
the IC's circuit to a ground wire coming from the parallel port,
otherwise those TTL lines from the parallel port would provide no
"signal" to the IC, correct?

As to how they share load, it depends on whether or not they are one
way sources (like transformer rectifier supplies) or two way (able to
deliver power or absorb it) as with batteries. But in both cases,
details count.

So hooking two batteries in series is different than hooking ac
adapters in series?


I'm confused on the whole ac adapter thing in general. Referring to
your typical generic unregulated wall-wart, they have a voltage and
milliamps rating. Does this mean the adapter has to be drawing the
rated amount of milliamps before it will deliver that particular
voltage? Or does it mean you have to have a load with a particular
amount of resistance before you get that number of volts?

I'm also curious if my speculation is correct on the resistance. Can
you figure out the internal resistance of an AC adapter using the
standard formula? Like using a 9v 500ma adapter, could you just do
9v/500ma = 18ohms? I guess I should actually ask if this 18 ohms is in
fact the internal resistance, or is this the amount of resistance your
circuit has to create before the adapter will generate the 9v? And if
the latter is the case, why would the adapter be rated in milliamps and
not resistance?

I keep hearing that it's okay to use higher-rated milliamp ac adapters
on devices if you don't have the device-rated amount available. But
wouldn't this result in higher voltage going into the device?

.



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